0
$\begingroup$

Let $k:\mathcal{O}\times\mathbb{R}^n\to\mathbb{R}$, with $\mathcal{O}\subset\mathbb{R}^m$ open, be such that $\forall x\in\mathcal{O}\quad k(x,\cdot)\in L^1(\mathbb{R}^n) $, i.e. the function $y\mapsto k(x,y)$ is Lebesgue summable on $\mathbb{R}^n$, according to the usual $n$-dimensional Lebesgue measure.

I read (theorem 1.d here, p. 2) that

if $\partial_{x_j}\partial_{x_i} k(\cdot,y)\in C(\mathcal{O})$ for almost all $y\in\mathbb{R}^n$ and there exists $ g\in L^1(\mathbb{R}^n)$ such that $$|\partial_{x_j}\partial_{x_i} k(x,y)|\le g(y)\quad\forall x\in\mathcal{O}\quad\text{for almost all }y\in\mathbb{R}^n,$$then$^1$ $$\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$where $x_i,x_j$ are components of $x$.

I see, by using the corollary to the dominated convergence theorem quoted in this question, that, provided that $\forall x\in\mathcal{O}\quad \partial_{x_i}k(x,\cdot)\in L^1(\mathbb{R}^n) $ (which I do not know how to prove$^2$), then$$\partial_{x_j}\int_{\mathbb{R}^n}\partial_{x_i}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$but then I do not know how we can "move" $\partial_{x_i}$ outside the integral.

How can we prove that $\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$? I thank any answerer very much.


Notes:

$^1$When I read a text I always assume that it is prefectly worded. Nevertheless, after many fruitless trials, I am not excluding that this result is intended to hold provided that the condition (b) here holds. I would be very grateful to any answerer confirming or denying that.

$^2$By using Fubini's theorem I only see that, once chosen an arbitrary interval $[a,t]$, the function $\partial_{x_i}k(x_t,\cdot)-\partial_{x_i}k(x_a,\cdot)$, where $x_t$ has $t$ as its $j$-th component, is summable and its integral is $\int_{[a,t]}\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y) \,d\mu_y \,d\mu_{x_j}$.

$\endgroup$
  • $\begingroup$ You may prove the statement by applying your corollary twice. In regards to your second footnote. The answer is no. In general, if you want derivatives to live in your $L^p$ space, you have to define them to exist. These are called Sobolev Spaces, usually denoted $W^{r,p}$ which means it's $L^p$ where all functions have $r$ derivatives that live in $L^p$ as well. $\endgroup$ – Jeb Mar 17 '16 at 15:02
  • $\begingroup$ @Jeb Thank you very much for your comment! In order to apply that corollary twice, I should know that $ \exists g\in L^1(\mathbb{R}^n)$ : for almost all $y\in\mathbb{R}^n$ and $\forall x\in\mathcal{O}\quad$ $|\partial_{x_i} k(x,y)|\le g(y)$, which I don't know how to prove... (my note 2 meant: can the fact that $k(x,\cdot)\in L^1(\mathbb{R}^n)$ together with the stated assumptions on $\partial_{x_j}\partial_{x_i} k(\cdot,y)$ be used to prove that?) $\endgroup$ – Self-teaching worker Mar 17 '16 at 15:19
0
$\begingroup$

I have been able to contact the author of the paper, who has told me that the assumption $$\exists G\in L^1(\mathbb{R}^n):|\partial_{x_i}k(x,y)|\le G(y),\quad\forall x\in\mathcal{O},\quad\text{for almost every }y\in\mathbb{R}^n$$made for (b) is implicitly made.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.